To determine which table represents a linear function, we need to check if the change in $y$ is consistent for equal changes in $x$. A function $y = mx + b$ is linear if the differences between consecutive $y$ values divided by the differences between the corresponding $x$ values (i.e., the slope) are constant.

Let's analyze each table:

Table 1

$x$ 1 1 1 1 1
$y$ −3 −2 −1 0 1

• The $x$ values are all the same, so we cannot determine a slope. This is not a function.

Table 2

$x$ −4 −2 0 2 4
$y$ 4 2 0 2 4

• Calculate the slope between each pair of points: $\text{Slope between } (−4, 4) \text{ and } (−2, 2) = \frac{2 - 4}{-2 - (-4)} = \frac{-2}{2} = -1$ $\text{Slope between } (−2, 2) \text{ and } (0, 0) = \frac{0 - 2}{0 - (-2)} = \frac{-2}{2} = -1$ $\text{Slope between } (0, 0) \text{ and } (2, 2) = \frac{2 - 0}{2 - 0} = \frac{2}{2} = 1$ $\text{Slope between } (2, 2) \text{ and } (4, 4) = \frac{4 - 2}{4 - 2} = \frac{2}{2} = 1$

• The slopes are not consistent, so this is not a linear function.

Table 3

$x$ −5 −3 −1 1 3
$y$ −1/2 1 2 7/2 5

• Calculate the slope between each pair of points: $\text{Slope between } (−5, -\frac{1}{2}) \text{ and } (−3, 1) = \frac{1 - (-\frac{1}{2})}{-3 - (-5)} = \frac{1 + \frac{1}{2}}{2} = \frac{\frac{3}{2}}{2} = \frac{3}{4}$ $\text{Slope between } (−3, 1) \text{ and } (−1, 2) = \frac{2 - 1}{-1 - (-3)} = \frac{1}{2} = \frac{1}{2}$ $\text{Slope between } (−1, 2) \text{ and } (1, \frac{7}{2}) = \frac{\frac{7}{2} - 2}{1 - (-1)} = \frac{\frac{7}{2} - \frac{4}{2}}{2} = \frac{\frac{3}{2}}{2} = \frac{3}{4}$ $\text{Slope between } (1, \frac{7}{2}) \text{ and } (3, 5) = \frac{5 - \frac{7}{2}}{3 - 1} = \frac{\frac{10}{2} - \frac{7}{2}}{2} = \frac{\frac{3}{2}}{2} = \frac{3}{4}$

• The slopes are consistent, so this is a linear function with slope $\frac{3}{4}$.

Table 4

$x$ −6 −4 −2 0 2
$y$ 5 $\frac{13}{3}$ $\frac{11}{3}$ 3 $\frac{7}{3}$

• Calculate the slope between each pair of points: $\text{Slope between } (−6, 5) \text{ and } (−4, \frac{13}{3}) = \frac{\frac{13}{3} - 5}{-4 - (-6)} = \frac{\frac{13}{3} - \frac{15}{3}}{2} = \frac{\frac{-2}{3}}{2} = \frac{-1}{3}$ $\text{Slope between } (−4, \frac{13}{3}) \text{ and } (−2, \frac{11}{3}) = \frac{\frac{11}{3} - \frac{13}{3}}{-2 - (-4)} = \frac{\frac{-2}{3}}{2} = \frac{-1}{3}$ $\text{Slope between } (−2, \frac{11}{3}) \text{ and } (0, 3) = \frac{3 - \frac{11}{3}}{0 - (-2)} = \frac{\frac{9}{3} - \frac{11}{3}}{2} = \frac{\frac{-2}{3}}{2} = \frac{-1}{3}$ $\text{Slope between } (0, 3) \text{ and } (2, \frac{7}{3}) = \frac{\frac{7}{3} - 3}{2 - 0} = \frac{\frac{7}{3} - \frac{9}{3}}{2} = \frac{\frac{-2}{3}}{2} = \frac{-1}{3}$

• The slopes are consistent, so this is a linear function with slope $\frac{-1}{3}$.

Conclusion

The tables that represent a linear function are Table 3 and Table 4.

Would you like more details or have any questions?

Here are some questions you can ask next:

1. How do you find the slope of a line given two points?
2. Can you explain the properties of linear functions?
3. What is the difference between a linear and a nonlinear function?
4. How do you graph a linear equation?
5. What is the significance of the y-intercept in a linear function?

Tip: To determine if a set of points forms a linear function, always check if the change in $y$ is proportional to the change in $x$ for all pairs of points.